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Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.
To do:
We have to prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.
Solution:
Let $x=\sqrt{3}+\sqrt{5}$ be a rational number.
Squaring on both sides, we get,
$x^{2}=(\sqrt{3}+\sqrt{5})^{2}$
$x^{2}=3+5+2 \times \sqrt{3} \times \sqrt{5}$
$x^{2}=8+2 \sqrt{15}$
$x^{2}-8=2 \sqrt{15}$
$\Rightarrow \frac{x^{2}-8}{2}=\sqrt{15}$
$x=\sqrt{3}+\sqrt{5}$ is a rational number.
This implies,
$\frac{x^{2}-8}{2}$ should be a rational number.
$\Rightarrow \sqrt{15}$ is a rational number.
But it is not possible as $\sqrt{15}$ is an irrational number.
Therefore, our assumption that $x=\sqrt{3}+\sqrt{5}$ is a rational number is false.
Hence, $\sqrt{3}+\sqrt{5}$ is an irrational number.
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